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Baby Analysis with Musings

Created
Last Edited byTThe Grand Inquisitor
Last Edited Time

0. 驯龙集

1. Limit of Sequence

Definitions

不论出身,不论生平,只论终相。然无终何以论终?身之棺具体,心之椁无形,可小之又小,小小不息。

  1. limnan=l\lim_{n\rarr\infty}a_n= l

    ε>0,N(ε),s.t. anl<ε,nN(ε).\forall \varepsilon>0,\exists N(\varepsilon),s.t.~|a_n-l|<\varepsilon,\forall n\ge N(\varepsilon). Back to index

    • Equivalently,
      1. (UF-1) What’s new: ...c......\le c...

        ε>0,N(ε),s.t. anlcε,nN(ε) and SOME constant c.\forall \varepsilon>0,\exists N(\varepsilon),s.t.~|a_n-l|\le c\cdot\varepsilon,\forall n\ge N(\varepsilon)\text{ and SOME constant }c.

        Proof of equivalence: From standard to UF-1, just take c=1c=1, the other way around, see here.

        Warning: ε,N(ε)\varepsilon,N(\varepsilon) cannot appear in cc !!! HW1P5 wrong original solution.

      1. (UF-2) What’s new: only talk about tiny teeny ε\varepsilon. anl|a_n-l| 明显有最大值时考虑使用。

        ε0>0 s.t. 0<ε<ε0, N(ε) s.t. anlcε xN(ε).\exists \varepsilon_0>0~s.t.~\forall0<\varepsilon<\varepsilon_0,~\exists N(\varepsilon)~s.t.~|a_n-l|\le c\varepsilon~\forall x\ge N(\varepsilon).

        Proof of equivalence: From UF-1 to UF-2, obvious cuz UF-2 only talks about a subset of UF-1’s cases. From UF-2 to UF-1, see here.

        • Example 1: Prove limn11n=1\lim_{n\rarr\infty}\sqrt{1-\frac1n}=1. Back to index

          WTS ε>0,111n<ε.{ε1always holdsε<1calculate Nε\forall \varepsilon>0,|1-\sqrt{1-\frac1n}|<\varepsilon. \begin{cases} \varepsilon\ge 1 & \text{always holds}\\ \varepsilon< 1 & \text{calculate }N_\varepsilon \end{cases}

      1. (Cauchy Sequence)

        ε>0,N(ε),s.t. anamcε,n,mN(ε).\forall \varepsilon>0,\exists N(\varepsilon),s.t.~|a_n-a_m|\le c\varepsilon,\forall n,m\ge N(\varepsilon).

        Proof of equivalence: \Rarr” here, \Larr” here.

        • Hw3 T2 Definition of {an}n=1\{a_n\}_{n=1}^\infin NOT Cauchy. Back to index

          ε0>0 s.t. N,m,n>N s.t. anamε0.\exists \varepsilon_0>0~s.t.~\forall N,\exists m,n>N~s.t.~|a_n-a_m|\ge\varepsilon_0.

  1. limnanl\lim_{n\rarr\infty}a_n\ne l

    naive: ε0>0,N,nN,s.t. anlε0.\exists\varepsilon_0>0,\forall N,\exists n\ge N,s.t.~|a_n-l|\ge\varepsilon_0.

    polished: ε0>0 and bad subseq {ank}k=1 s.t. anklε0,k1.\exists \varepsilon_0>0\text{ and bad subseq }\{a_{n_k}\}^\infin_{k=1}~s.t.~|{a_{n_k}-l}|\ge\varepsilon_0,\forall k\ge1. Back to index

    We now show the construction process of n1,n2,..,nk,...n _1,n_2,..,n_k,... use the \forall term

    Note that he key is that both nkn_k and kk goes to \infin (partly by definition of subsequence).

    Take N1=1N_1=1, n(N1)N1,s.t.\exists n(N_1)\geq N_1,s.t. an(N1)lϵ0|a_{n(N_1)}-l| \geq \epsilon_0.

    Relabel n(N1)n(N_1) as n1n_1.

    To guarantee n2>n1n_2 > n_1, we take N2=n1+1N_2=n_1+1, then there exist n(N2)N2n(N_2) \geq N_2 such that an(N2)lϵ0|a_{n(N_2)}-l| \geq \epsilon_0. Relabel n(N2)n(N_2) as n2n_2 and we know n2N2>n1n_2 \geq N_2 >n_1.

    Repeat this process, we have bad subseq {ank}k=1\{a_{n_k}\}^\infin_{k=1}.

  1. limnan=\lim_{n\rarr\infty}a_n= \infty

    M>0,N(M) s.t. an>M nN(M).\forall M>0,\exists N(M)~s.t.~a_n>M~\forall n\ge N(M).

  1. limnan\nexists\lim_{n\rarr\infty}a_n

    第一个是极限存不存在的问题,等于一个 finite number ll 或者等于±\pm \infty,都算是极限存在。极限不存在使用liman\lim a_n 就不是well-defined,所以我觉得l,limanl\forall l, \lim a_n \neq l也不是一个很严谨的说法,因为极限不存在的话是不能用liman\lim a_n这个符号的(就这样说有点默认{an}\{a_n\}的极限存在了,只是不等于ll罢了)。在(1)n(-1)^n的例子里,说它极限不存在我们也没有严谨证明。第二个问题是极限存在后,这个极限是不是有限的。

    ϵN\epsilon-N 这种定义是把 是否convergent和converge到哪里混合在一起的,cauchy的定义就只涉及convergent,但不说converge到哪里。所以我想到的极限不存在的说法是 1. 不是柯西的 (这样它是divergent)+ 2. 不趋于±\pm \infty.

    行吧,应该不是big issue. 估计这几个terminologies都没有很统一,符号的意思大概比语言的意思更本质。

Basic Facts
  • 1.1 收敛必有界 {an} is convergent{an} is bounded\{a_n\} \text{ is convergent}\Rarr \{a_n\}\text{ is bounded}

    断点法首秀。点之前利用finite # of #取max,点之后利用极限的定义和triangle inequality.

    Let liman=l take ε=411,N(411) s.t. anlε=411,nN(411)anlanl411an411+l,nN(411).\lim a_n=l\Rarr\text{ take }\varepsilon=411,\exists N(411) ~s.t.~|a_n-l|\le\varepsilon=411,\forall n\ge N(411)\Rarr|a_n|-|l|\le|a_n-l|\le411\Rarr|a_n|\le411+|l|,\forall n\ge N(411).

    As for n<N(411), take M=max{411+l,a1,a2,...,aN(411)1}anM,n1.n<N(411),\text{ take }M=max\{411+|l|,|a_1|,|a_2|,...,|a_{N(411)-1}|\}\Rarr|a_n|\le M, \forall n\ge 1.\blacksquare

    Remark: {an} is bounded{an} is convergent\{a_n\} \text{ is bounded}\nRightarrow \{a_n\}\text{ is convergent}

  • 1.2 Squeezing Theorem: Suppose anbncn & liman=limcn=llimbn exists and=la_n\le b_n\le c_n~\&~\lim a_n=\lim c_n=l\Rarr\lim b_n\text{ exists and}=l

    只要足够大……那什么才叫足够大?\Rarrmax{N1,N2,...}\max\{N_1,N_2,...\}.

    Proof: liman=limcn=lan,cn(lε,l+ε),nmax{Na(ε),Nc(ε)}bn(lε,l+ε),nmax{Na(ε),Nc(ε)}.\because\lim a_n=\lim c_n=l\therefore a_n,c_n\in(l-\varepsilon,l+\varepsilon),\forall n\ge \max\{N_a(\varepsilon),N_c(\varepsilon)\}\Rarr b_n\in(l-\varepsilon,l+\varepsilon),\forall n\ge \max\{N_a(\varepsilon),N_c(\varepsilon)\}.\blacksquare

  • 1.3 四则运算:Suppose liman=a,limbn=b\lim a_n=a,\lim b_n=b
    • lim(an±bn)=liman±limbn\lim (a_n\pm b_n)=\lim a_n\pm\lim b_n (证明思路同1.2)
    • lim(can)=climan\lim(ca_n)=c\lim a_n
    • limanbn=limanlimbn\lim a_nb_n=\lim a_n\lim b_n

      Proof hint: anbnab=(ana)(bnb)+a(bnb)+b(ana)a_nb_n-ab=(a_n-a)(b_n-b)+a(b_n-b)+b(a_n-a)

    • limanbn=limanlimbn, if limbn0\lim \frac{a_n}{b_n}=\frac{\lim a_n}{\lim b_n},\text{ if }\lim b_n\ne 0

      Proof: see here.

  • 1.4 极限与序列元素相对大小一致: Suppose liman=a,limbn=b,anbn\lim a_n=a,\lim b_n=b,a_n\le b_n for n some N, then ab.n\ge\text{ some }N,\text{ then }a\le b.

    Proof: (By contradiction) Suppose a>ba>b, we try to create a circumstance under which an>bna_n>b_n for large enough n.n.

    ε>0,N s.t. aε<an<a+ε,bε<bn<b+ε,n>N\forall \varepsilon>0, \exists N~s.t.~a-\varepsilon<a_n<a+\varepsilon,b-\varepsilon<b_n<b+\varepsilon,\forall n>N. Dream: b+ε<aεε<ab2.b+\varepsilon<a-\varepsilon\Rarr\varepsilon<\frac{a-b}2. So we take ε=ab411, then an<a+ε<bε<bn if nN(ab411).\varepsilon= \frac{a-b}{411},\text{ then }a_n<a+\varepsilon<b-\varepsilon<b_n\text{ if }n\ge N(\frac{a-b}{411}). \blacksquare

    Corollary:

    If anm,nN, then if liman,limanm.a_n\ge m,\forall n\ge N,\text{ then if }\exists\lim a_n,\lim a_n\ge m.

    However, "an>m,nN, then if liman,liman>m""a_n> m,\forall n\ge N,\text{ then if }\exists\lim a_n,\lim a_n> m" is WRONG.

  • 1.5 子列与原列极限一致 Suppose liman=l,\lim a_n=l, Then for any subsequence {ank}k=1\{a_{n_k}\}^\infty_{k=1}, we have limank=l.\lim a_{n_k}=l.

    Proof see here.

  • 1.6 原列与子子列极限一致 Suppose {an}\{a_n\} satisfies: \forall subsequence of {an}\{ a_n\}, \exist  sub-subsequence {ank}k=1\{a_{n_k}\}_{k=1}^{\infty}, s.t. limkank\lim_{k\to \infty} a_{n_k} exists &=l=l , where ll is independent of the choice of subsequence. Then limkan\lim_{k\to \infty} a_{n} exists &=l=l .

    Proof see here.

    Remark: by 1.6, the converse of 1.5 holds: if for any subsequence {ank}k=1\{a_{n_k}\}^\infty_{k=1}, we have limank=l,\lim a_{n_k}=l, then liman=l.\lim a_n=l.

    1.6 is more general than the converse of 1.5, because it only requires each subsequence to contain one (\exists) subsubseq that converge to ll, whereas the converse of 1.5 says any subseq itself should converge to ll. Note that \exist  sub-subsequence {ank}k=1\{a_{n_k}\}_{k=1}^{\infty}, s.t. limkank\lim_{k\to \infty} a_{n_k} exists &=l=l , doesn’t necessarily give limit of subseq also exists and is ll  (e.g., (1)n(-1)^n). However, if we include the \forall part into our consideration, is 1.6 really more general than the converse of 1.5? Can we find a seq where “ \forall subsequence of {an}\{ a_n\}, \exist  sub-subsequence {ank}k=1\{a_{n_k}\}_{k=1}^{\infty}, s.t. limkank\lim_{k\to \infty} a_{n_k} exists &=l=l , where ll is independent of the choice of subsequence” is satisfied yet “\forall  subsequence {ank}k=1\{a_{n_k}\}^\infty_{k=1}, we have limank=l\lim a_{n_k}=l” is not? @The Grand Inquisitor

    @Cindy Zeng 我觉得没有,因为如果前面的条件满足,那么liman\lim a_n exists & =l=l, 然后由fact 5就可以推出后面那个条件。 所以这两个条件可以互相imply,它们应该是等价的。

    @The Grand Inquisitor 我也觉得没有更general…

  • (1.7) Monotone Convergence Theorem

    Suppose {an}\{a_n\} is monotone & bounded, then limnan  as a finite #.\lim_{n\rarr\infty}a_n~\exists\text{ as a finite } \#.

    • Supremum and Infimum
      • A real # ss is said to be the “least upper bound (supremum 上确界)” of SS if
        1. ss is an upper bound of SS, i.e., xS,xs.\forall x\in S, x\le s.
        1. ε>0,sε<xε for some xεS.\forall \varepsilon>0,s-\varepsilon<x_\varepsilon \text{ for some }x_\varepsilon\in S.

        Intuitively, either s=maxSs=\max S or elements of SSss 之下紧密排布,“没有最近,总有更近的。”

      • A real # ss is said to be the “greates lower bound (infimum 下确界)” of SS if
        1. ss is an lower bound of SS, i.e., xS,xs.\forall x\in S, x\ge s.
        1. ε>0,s+ε>xε for some xεS.\forall \varepsilon>0,s+\varepsilon>x_\varepsilon \text{ for some }x_\varepsilon\in S.

      Remark: \forall upper bound MM of S,S, we have MsupSM\ge\sup S; \forall lower bound mm of S,S, we have minfSm\le\inf S.

      • Example: Hw2 T1 Back to index

        (iv) rational number: can be represented by pq\frac pq where p, q are integers. 思路和证明极限等于某某差不多。

    • The Least Upper Bound Axiom (公理)

      If SS  is bounded from above, then supS\sup S exists. 总存在最小的上界。

      \rarr Theorem: If SS  is bounded from below, then infS\inf S exists.

      Proof: Let T=STT=-S\Rarr T is bounded from above. By axiom, supT\sup T  exists as a finite number, and it satisfies

      1. sT, we have ssupTssupT,sS\forall -s\in T,\text{ we have }-s\le \sup T\Rarr s\ge-\sup T,\forall s\in S
      1. ε>0,sεT s.t. supTε<sεsε<supT+εsupT is infS.\forall \varepsilon>0,\exists -s_\varepsilon\in T~s.t.~\sup T-\varepsilon<-s_\varepsilon\Rarr s_\varepsilon<-\sup T+\varepsilon\Rarr -\sup T \text{ is }\inf S.

      By-product: sup(S)=infS;sup(S)=inf(S)\sup(-S)=-\inf S;-\sup(S)=\inf (-S) 名字相对,负号两次

    • Proof of MCT

      Only consider the case of increasing ana_n. {an}\{a_n\} is bounded from above Axiomsup{an}n=1\overset{\text{Axiom}}{\Rarr}\sup\{a_n\}_{n=1}^\infty exists, we let it be ll. By definition:

      1. anl,n a_n\le l,\forall n~\bigstar
      1. ε>0,lε<aNε for some Nε.\forall \varepsilon>0,l-\varepsilon<a_{N_\varepsilon}\text{ for some }N_\varepsilon.  an lε<aNεanl<l+ε,nNεanl<εliman=l.a_n\uarr~\Rarr l-\varepsilon<a_{N_\varepsilon}\le a_n\overset{\bigstar}{\le}l<l+\varepsilon,\forall n\ge N_\varepsilon\Lrarr |a_n-l|<\varepsilon\Rarr \lim a_n=l. \blacksquare 会给挤压到天花板上。

      Remark: If ana_n\uarr & not bounded from above, liman\lim a_n\rarr\infty (you’re going to heaven); If ana_n\darr & not bounded from below, liman\lim a_n\rarr-\infty (you’re going to hell). Meaning if we adopt the conceptual framework described in the beginning of this note, monotonicity \Rarr limit must exist (either as a finite number or as infinities).

    • Example

      Example 4: Does limn(1+1n)n\lim_{n\to \infty} (1+\frac{1}{n})^n exist? (The definition of ee; proved by M.C.T.)

      Solution see here.

2. Upper and Lower Limits

Motivation

limnan\lim_{n\rarr\infty} a_n does not always exist, but upper and lower limites always exist. And we can use them to prove the existence of / directly find limnan\lim_{n\rarr\infty} a_n. Prof. Wang loves it. Powerful stuff.

Definition

Consider {an}n=n0\{a_n\}^\infty_{n=n_0}, WLOG, assume n0=1n_0=1 for convenience (we don’t really care about n0n_0).

y1=inf{a1,a2,...,an,...} ± allowed.y2=inf{a2,a3,...,an,...}yn=inf{an,...}y_1=\inf\{a_1,a_2,...,a_n,...\}~^*\pm\infty\text{ allowed.}\\ y_2=\inf\{a_2,a_3,...,a_n,...\}\\y_n=\inf\{a_n,...\}

And in this way we construct a new sequence {yn}n=1\{y_n\}_{n=1}^\infty and yny_n\uarr. By MCT, limnyn\lim_{n\rarr\infty} y_n exists and may be ±\pm\infty. We call limnyn\lim_{n\rarr\infty} y_n the “lower limit” of {an}\{a_n\}, and we denote it as limninfan\lim_{n\rarr\infty}\inf a_n or limnan\varliminf_{n\rarr\infty} a_n.

Similarly, we can construct a new sequence {zn}n=1\{z_n\}_{n=1}^\infty and znz_n\darr, where zn=sup{an,an+1,...}z_n=\sup\{a_n,a_{n+1,...}\}. We call limnzn\lim_{n\rarr\infty} z_n the “upper limit” of {an}\{a_n\}, and we denote it as limnsupan\lim_{n\rarr\infty}\sup a_n or limnan\varlimsup_{n\rarr\infty} a_n.

Basic Facts
  • 2.1 下极限小于上极限 limnanlimnan\varliminf_{n\rarr\infty} a_n\le \varlimsup_{n\rarr\infty} a_n

    Proof: ynznF1.4limynlimzn.y_n\le z_n\overset{F1.4}{\Rarr}\lim y_n\le \lim z_n. \blacksquare

  • 2.2 中原存则蛮夷附之 limnan\lim_{n\rarr\infty} a_n exists (allow ±\pm\infty) limnan=limnan(=limnan)\Lrarr\varliminf_{n\rarr\infty} a_n=\varlimsup_{n\rarr\infty} a_n(=\lim_{n\rarr\infty} a_n)

    Proof:

    \Rarr”.

    case 1: limnan=l\lim_{n\rarr\infty} a_n=l (finite #) ε>0,N s.t. lε<an<l+ε,nNlεynznl+ε,nNlimyn=l=limzn.\Rarr\forall\varepsilon>0,\exists N~s.t.~l-\varepsilon<a_n<l+\varepsilon,\forall n\ge N\Rarr l-\varepsilon\le y_n\le z_n\le l+\varepsilon,\forall n\ge N\Rarr \lim y_n=l=\lim z_n.

    case 2: limnan=M>0,N s.t. anM,nNynMlimnyn=F2.1limnan=limnan=\lim_{n\rarr\infty} a_n=\infin\Rarr \forall M>0,\exists N~s.t.~a_n\ge M,\forall n\ge N\Rarr y_n\ge M\Rarr \lim_{n\rarr\infty} y_n=\infty\overset{F2.1}{\Rarr}\varliminf_{n\rarr\infty} a_n= \varlimsup_{n\rarr\infty} a_n=\infty

    case 3: -\infty. Similar to case 2. \blacksquare

    \Larr”.

    case 1: limnan=l\lim_{n\rarr\infty} a_n=l (finite #) ε>0,N s.t. lε<yn,zn<l+ε,nNlεynanznl+ε,nNliman=l.\Rarr\forall\varepsilon>0,\exists N~s.t.~l-\varepsilon<y_n,z_n<l+\varepsilon,\forall n\ge N\Rarr l-\varepsilon\le y_n\le a_n\le z_n\le l+\varepsilon,\forall n\ge N\Rarr \lim a_n=l.

    case 2: limnyn=M>0,N s.t. ynM,nNanMlimnan=.\lim_{n\rarr\infty} y_n=\infin\Rarr \forall M>0,\exists N~s.t.~y_n\ge M,\forall n\ge N\Rarr a_n\ge M\Rarr \lim_{n\rarr\infty} a_n=\infty.

    case 3: -\infty. Similar to case 2. \blacksquare

  • 2.3.1 上下极限之外,远则不遇 If a0>limnan, then N s.t. a0>an,nN        If b0<limnan, then N s.t. b0<an,nN\text{If }a_0>\varlimsup_{n\rarr\infty} a_n,\text{ then }\exists N~s.t.~a_0>a_n,\forall n\ge N\\~~~~~~~~\text{If }b_0<\varliminf_{n\rarr\infty} a_n,\text{ then }\exists N~s.t.~b_0<a_n,\forall n\ge N
    2.3.2
    总存在收敛到上极限/下极限的子序列 Define a set E={a subseq ank with limit a, allow a to be ±.} Then limnan,limnanEE=\{a|\exists \text{ subseq }a_{n_k}\text{ with limit } a,\text{ allow }a\text{ to be }\pm\infty.\}\text{ Then }\varliminf_{n\rarr\infty} a_n, \varlimsup_{n\rarr\infty} a_n\in E.
    2.3.3
    (上)下极限是E的(上)下确界 limnan=infE; limnan=supE.\varliminf_{n\rarr\infty} a_n=\inf E;~\varlimsup_{n\rarr\infty} a_n=\sup E.

    Proofs:

    2.3.1

    limnzn=lˉε>0,N s.t. if nN,anzn<lˉ+ε. Take ε=a0lˉan<a0,nN.\lim_{n\rarr\infty} z_n=\bar l\Rarr\forall\varepsilon>0,\exists N~s.t.~\text{if }n\ge N,a_n\le z_n<\bar l+\varepsilon.\text{ Take } \varepsilon=a_0-\bar l\Rarr a_n<a_0,\forall n\ge N.

    Rest similar.

    2.3.2

    See here.

    2.3.3

    By Fact 2.3.2 we know that limnanElimnansupE. WTS aE,limnana().\varlimsup_{n\rarr\infty} a_n\in E\Rarr\varlimsup_{n\rarr\infty} a_n\le\sup E.\text{ WTS }\forall a\in E, \varlimsup_{n\rarr\infty} a_n\ge a (\bigstar). By definition of E, subseq {ank}k=1 s.t. anka.E,\exists \text{ subseq }\{a_{n_k}\}_{k=1}^\infty~s.t.~a_{n_k}\rarr a. limzn=lˉlimznk=lˉ. znkankF1.4lˉa,aElˉsupEl=supE.\because \lim z_n=\bar l\therefore\lim z_{n_k}=\bar l.~z_{n_k}\ge a_{n_k}\overset{F1.4}{\Rarr}\bar l\ge a,\forall a\in E\Rarr \bar l\ge \sup E\overset{\bigstar}{\Rarr}l= \sup E. \blacksquare

  • 2.4 上下极限与序列元素相对大小一致(类比F1.4) If anbn,n some N, then limnanlimnbn,limnanlimnbn.\text{If }a_n\le b_n,\forall n\ge\text{ some }N,\text{ then }\varliminf_{n\rarr\infty} a_n\le \varliminf_{n\rarr\infty} b_n,\varlimsup_{n\rarr\infty} a_n\le \varlimsup_{n\rarr\infty} b_n.

    Proof: @The Grand Inquisitor 小天,帮我看看这个证明对不对😭😫

    Case 1: limnbn\varliminf_{n\rarr\infty} b_n is a finite number.

    Suppose limnan>limnbn\varliminf_{n\rarr\infty} a_n> \varliminf_{n\rarr\infty} b_n, by Fact 2.3.1, N1 s.t. limnynb<an,nN1\exists N_1~s.t.~\lim_{n\rarr\infty} y^b_n<a_n,\forall n\ge N_1, take ϵ=aN1limnynb2, then yN1b<limnynb+ϵ<aN1\epsilon=\frac{a_{N_1}-\lim_{n\rarr\infty} y^b_n}2,\text{ then }y_{N_1}^b<\lim_{n\rarr\infty} y^b_n+\epsilon<a_{N_1}. Let N2=N1+1,ϵ2=...N_2=N_1+1,\epsilon_2=... we can construct a bad sequence of {yNnb}n=1\{y^b_{N_n}\}_{n=1}^\infty where yNnb<aNn,n1,y^b_{N_n}<a_{N_n},\forall n\ge 1, contradicting anbn,n some N.a_n\le b_n,\forall n\ge\text{ some }N. 到红色为止就可以结束了,前面加一个bN1<yN1bb_{N_1}<y_{N_1}^b, 就能得出bN1<aN1b_{N_1}<a_{N_1}的矛盾 (这个只是在某个值上面满足,似乎不够)

    我当时想了想好像不行,我们需要一个完整的序列来contradict n\forall n\ge some N 那个部分,因为N1N_1 不行,或许比他大的some N 还是可以work的。嗯你说的对!本来以为真就是F1.4的翻版,一试发现似乎没那么简单O.o#.

    我觉得利用ynaynby_n^a\leq y_n^b (集合中每个元素都更小,inf也更小),让然后就可以直接用fact 1.4 说 limynalimynb\lim y_n^a\leq \lim y_n^b, 也即 limanlimbn\underline{\lim} a_n \leq \underline{\lim} b_n @The Grand Inquisitor 天,我突然想到一个问题,我们并不知道 limyna,limynb\lim y_n^a, \lim y_n^b 存在啊?(@Cindy Zeng 根据MCT,limyn\lim y_nlimzn\lim z_n 都存在,所以说liminf 和 limsup总是有的)

    诶,有道理。别走 @The Grand Inquisitor 可是那我那个复杂的方法对不对啊,感觉每次取一个不同的ϵ\epsilon很心虚。上次好像就错过一次,因为不能反复取ϵ\epsilon.

    这里没问题,极限不存在和不cauchy的例子不能反复取不同的ε\varepsilon是因为都是针对的同一个bad ε\varepsilon,但像证明有一个子序列极限为limsup的时候,第n轮的εn=1n\varepsilon_n=\frac{1}{n},每轮都是递减的。

    嗯对,有时可以,有时不行。可能就要注意分清什么时候要一ε\varepsilon到底,什么时候只需要找到一个它就好了(可能当ε\forall \varepsilon是条件的一部分的时候,找到一个就行,但当我们需要说明ε0\exists\varepsilon_0 balabala 的时候就必须一ε\varepsilon到底)。主要是之前我错的那个地方我也记不太清了呜呜,好像是某道作业题,没事,应该找得到。

  • 2.5 (Useful trick) liman=llimnanl=0.\lim a_n=l\Lrarr\varlimsup_{n\rarr\infty} |a_n-l|=0. 很有用!尤其是当我们不知道lim是否存在的时候。

    0limanllimanl=0limanl=0.0\le \varliminf|a_n-l|\le \varlimsup|a_n-l|=0\Rarr\lim|a_n-l|=0.

  • 2.6 合租房比整租房扁 limn(an+bn)limnan+limnbn     limn(an+bn)limnan+limnbn\varlimsup_{n\rarr\infty} (a_n+b_n)\le\varlimsup_{n\rarr\infty} a_n+\varlimsup_{n\rarr\infty}b_n\\~~~~~\varliminf_{n\rarr\infty} (a_n+b_n)\ge\varliminf_{n\rarr\infty} a_n+\varliminf_{n\rarr\infty}b_n (True as long as \infin-\infin does not appear on RHS.)

    Proof (Hw2T4.2, 4.3): Back to index

    规整和上确界≤打乱和上确界≤分开上确界(让我想起高中作文素材曾摘了一个形容独立性的重要性的美句哈哈哈,!找到了!“廊柱分立才能撑起庙宇。橡树和松柏也不能在彼此的阴影中生长。”——纪伯伦《先知》顺便看到了一句无关的,分享分享:“有人曾问过王劲松,荀彧为什么会相信曹操可以帮他匡扶汉室?王劲松用剧中的一句词作答:万古长夜当中,哪怕有一盏微弱的光芒,都会让你身不由己追随,至死方休。” @The Grand Inquisitor )军师联盟😭😭😭其实这个(包括下面那个)的证明思路有点像minmax。什么minmax? 为什么我第一反应是信息竞赛里的什么东西???证明零和博弈纳什均衡的一个定理

    我不知道这个😓,但是我第一眼看到你这个证明的时候就觉得这个思路很有前途😁。

    Let An={an,an+1,...},Bn={bn,bn+1,...},Sn={an+bn,an+1+bn+1,...},SSn={ai+bjin,jn,i,jZ}.ai+bjSSn,ai+bjsup(An)+sup(Bn)sup(SSn)sup(An)+sup(Bn).A_n=\{a_n,a_{n+1},...\},B_n=\{b_n,b_{n+1},...\},S_n=\{a_n+b_n,a_{n+1}+b_{n+1},...\},SS_n=\{a_i+b_j|i\ge n,j\ge n,i,j\in\Z\}. \forall a_i+b_j\in SS_n,a_i+b_j\le \sup(A_n)+\sup(B_n)\Rarr\sup(SS_n)\le \sup(A_n)+\sup(B_n).

    SnSSnai+biSn,ai+biSSnai+bisup(SSn)sup(Sn)sup(SSn)sup(Sn)sup(An)+sup(Bn)limn(an+bn)limnan+limnbn\because S_n\sub SS_n\therefore\forall a_i+b_i\in S_n,a_i+b_i\in SS_n \therefore a_i+b_i\le\sup(SS_n)\Rarr\sup(S_n)\le \sup(SS_n)\Rarr\sup(S_n)\le \sup(A_n)+\sup(B_n)\Rarr\varlimsup_{n\rarr\infty} (a_n+b_n)\le\varlimsup_{n\rarr\infty} a_n+\varlimsup_{n\rarr\infty}b_n.

  • 2.7 脱帽无妨 If limnan\lim_{n\rarr\infty} a_n exists as a finite #,

    limn(an+bn)=limnan+limnbnlimn(an+bn)=limnan+limnbn\varlimsup_{n\rarr\infty} (a_n+b_n)=\lim_{n\rarr\infty} a_n+\varlimsup_{n\rarr\infty}b_n\\\varliminf_{n\rarr\infty} (a_n+b_n)=\lim_{n\rarr\infty} a_n+\varliminf_{n\rarr\infty}b_n

    Proof (Hw2T4.4): Back to index

    还是关于上下极限和极限三者分析的老思路:筷子都夹住了,中间必有汤圆。

    Preserve the notations used in proving Fact 2.6. liman=liman=liman ().\lim a_n=\varliminf a_n=\varlimsup a_n~(\star). ai+biSn,i.e.,in, sup(Sn)ai+biinf(An)+sup(Bn)inf(An)+sup(Bn)sup(Sn)sup(An)+sup(Bn)limnan+limnbnlimn(an+bn)limnan+limnbnlimn(an+bn)=limnan+limnbn\forall a_i+b_i\in S_n,i.e.,i\ge n,~\sup(S_n)\ge a_i+b_i\ge\inf(A_n)+\sup(B_n)\Rarr \inf(A_n)+\sup(B_n)\le \sup(S_n)\le \sup(A_n)+\sup(B_n)\Rarr \varliminf_{n\rarr\infty} a_n+\varlimsup_{n\rarr\infty} b_n\le \varlimsup_{n\rarr\infty} (a_n+b_n)\le \varlimsup_{n\rarr\infty} a_n+\varlimsup_{n\rarr\infty}b_n\overset{\star}{\Rarr}\varlimsup_{n\rarr\infty} (a_n+b_n)= \lim_{n\rarr\infty} a_n+\varlimsup_{n\rarr\infty}b_n.

  • 2.8 首尾颠倒,负号两次 limn(an)=limnan,limn(an)=limnan\varlimsup_{n\rarr\infty} (-a_n)=-\varliminf_{n\rarr\infty} a_n, \varliminf_{n\rarr\infty} (-a_n)=-\varlimsup_{n\rarr\infty} a_n

    Proof (Hw2T4.1): Back to index

    Let Sn={an,an+1,...},Sn={an,an+1,...},zn=sup(Sn),yn=inf(Sn)S_n=\{a_n,a_{n+1},...\},-S_n=\{-a_n,-a_{n+1},...\},z_n=\sup(-S_n),y_n=\inf(S_n), by the byproduct of the least upper bound axiom, zn=ynlimn(an)=limnzn=limnyn=limnanz_n=-y_n\Rarr\varlimsup_{n\rarr\infty}(-a_n)=\lim_{n\rarr\infty} z_n=-\lim_{n\rarr\infty} y_n=-\varliminf_{n\rarr\infty} a_n.

  • Theorem: Bolzano-Weierstrass Theorem

    If {an}\{a_n\} is bounded, then it contains a convergent subseq.

    Proof:

    {an}\{a_n\} is bounded liman\Rarr\varlimsup a_n is finite. limanE\varlimsup a_n\in E\Rarr\exists a subseq converging to liman\varlimsup a_n.

3. Series

Definition: Given a sequence of {an}\{a_n\}, we call
  • n=n0an\sum_{n=n_0}^{\infty} a_n an infinite series
  • sn=k=n0naks_n=\sum_{k=n_0}^{n} a_k partial sums of the sequence
  • {sn}\{s_n\} constitutes a new sequence: if limnsn\lim_{n\to \infty} s_n exists as a finite number, then we say n=n0an\sum_{n=n_0}^{\infty} a_n converges and define n=n0an=limnsn\sum_{n=n_0}^{\infty} a_n=\lim_{n\to \infty} s_n. Otherwise, we say n=n0an\sum_{n=n_0}^{\infty} a_n diverges.
Basic Facts
  • 3.0 开端不影响结局,殊始同归 n=1an\sum_{n=1}^{\infty} a_n and n=n0an\sum_{n=n_0}^{\infty} a_n have the same convergence/divergence nature. (although the value may be different).
  • 3.1 Divergence Test\colorbox{yellow}{Divergence Test} 最后增量需趋于零,天下归心: If an\sum a_n converges, then limnan=0\lim_{n \to \infty} a_n=0. Consequently, if limnan0\lim_{n \to \infty} a_n \neq 0. then an\sum a_n diverges. ( limnan=0\lim_{n \to \infty} a_n=0an\sum a_n converge 的必要不充分条件).

    Proof: Observe an=snsn1a_n=s_n-s_{n-1}. Take limit on both sides:

    liman=lim(snsn1)=limsnlimsn1=0\lim a_n=\lim (s_n-s_{n-1})=\lim s_n-\lim s_{n-1}=0.

  • 3.2 只要足够远,绝对片面和就足够小 an\sum a_n converges \Leftrightarrow ϵ>0,N,s.t.\forall \epsilon>0, \exists N, s.t.  n,mNn,m \geq N  (WLOG, nmn\geq m), k=mnak<ϵ|\sum_{k=m}^n a_k|< \epsilon.
    • Proof: an\sum a_n converges limsn\Leftrightarrow \lim s_n exists & finite Cauchy’s Criterion{sn}\overset{\text{Cauchy's Criterion}}{\Leftrightarrow} \{s_n\} is Cauchy.

      ϵ>0,N,s.t.n,mN,snsm1=k=mnak<ϵ\forall \epsilon>0, \exist N, s.t. \forall n,m \geq N, | s_n-s_{m-1}|=| \sum_{k=m}^{n} a_k |< \epsilon

    • Remark 整条尾巴都要足够扁 remainder shrinks to 0: If an\sum a_n converges, then n=kan0\sum_{n=k}^{\infty} a_n \to 0 as kk \to \infty.

      Proof: n=kak=n=1ansk1\sum_{n=k}^{\infty} a_k= \sum_{n=1}^{\infty} a_n-s_{k-1}. Take limits on both sides of the equation

      limkn=kak=n=1anlimksk1=limnsnlimnsn=0\lim_{k \to \infty} \sum_{n=k}^{\infty} a_k =\sum_{n=1}^{\infty} a_n-\lim_{k\to \infty} s_{k-1}=\lim_{n \to \infty} s_n-\lim_{n\to \infty }s_n=0

      Because n=1an=limn\sum_{n=1}^{\infty} a_n=\lim_{n \to \infty} is just a finite number

    • Example: 尾巴扁不下来
      • Hw3 T10(b): Suppose an>0a_n>0, Sn=a1+...+anS_n=a_1+...+a_n and an\sum a_n diverges. Prove that aN+1SN+1+...+aN+kSN+k1SNSN+k\frac{a_{N+1}}{S_{N+1}}+...+\frac{a_{N+k}}{S_{N+k}}\geq 1-\frac{S_N}{S_{N+k}} and deduce that ansn\sum \frac{a_n}{s_n} diverges. Back to index

        Note that

        1SNSN+k=SN+kSNSN+k=aN+1+...+aN+kSN+kaN+1SN+1+...+aN+kSN+k1-\frac{S_N}{S_{N+k}}=\frac{S_{N+k}-S_N}{S_{N+k}}=\frac{a_{N+1}+...+a_{N+k}}{S_{N+k}} \leq \frac{a_{N+1}}{S_{N+1}}+...+\frac{a_{N+k}}{S_{N+k}}

        The last inequality holds because an>0a_n>0 hence sns_n is increasing.

        To prove that anSn\sum \frac{a_n}{S_n} diverges, we verify the opposite of its convergent conditions

        Take ε=12\varepsilon=\frac{1}{2}, then N\forall N, we take n=N+1,m=N+kn=N+1, m=N+k, so that

        j=N+1N+kaj=aN+1SN+1+...+aN+kSN+k1SNSN+k|\sum_{j=N+1}^{N+k} a_j|=|\frac{a_{N+1}}{S_{N+1}}+...+\frac{a_{N+k}}{S_{N+k}}| \geq 1-\frac{S_N}{S_{N+k}}

        Since limSn=\lim S_n =\infty. For any fixed NN, we have limkSNSN+k=0\lim_{k\to \infty} \frac{S_N}{S_{N+k}}=0, so we can find a kk big enough, such that SNSN+k<12\frac{S_N}{S_{N+k}}<\frac{1}{2}. This means j=N+1N+kaj12|\sum_{j=N+1}^{N+k}a_j|\geq \frac{1}{2}.

      • Hw3 T11(a): Back to index

        amrm+...+anrn>am+...+anrm=rmrn+1rm=rmrn+rnrn+1rm=rmrn+anrm>rmrnrm=1rnrm.\frac{a_m}{r_m}+...+\frac{a_n}{r_n}>\frac{a_m+...+a_n}{r_m}=\frac{r_m-r_{n+1}}{r_m}=\frac{r_m-\colorbox{greenyellow}{$r_n+r_n$}-r_{n+1}}{r_m}=\frac{r_m-r_n+a_n}{r_m}>\frac{r_m-r_n}{r_m}=1-\frac{r_n}{r_m}. Since an\sum a_n converges, by remark of F3.2, rnr_n\rarr\infty as nn\rarr\infty m,n0>m s.t. rn0rm2amrm+...+an0rn0>1rn0rm12anrn\Rarr\forall m,\exists n_0>m~s.t.~r_{n_0}\le \frac{r_m}2\Rarr\frac{a_m}{r_m}+...+\frac{a_{n_0}}{r_{n_0}}>1-\frac{r_{n_0}}{r_m}\ge \frac12\Rarr\sum\frac{a_n}{r_n} is not Cauchy, therefore it diverges.

        For Cauchy stuff, you can tinker with m,nm,n since they are behind .\forall.

  • 3.3 Absolute Convergence 左边更难conv(所以更强),因为取和的话绝对值消除了正负相抵的可能性: n=1an\sum_{n=1}^{\infty} |a_n| converges \Rightarrow n=1an\sum_{n=1}^{\infty} a_n converges.

    Proof: By Fact 2, ϵ>0,N,s.t.n,mN,\forall \epsilon>0, \exist N, s.t. \forall n,m\geq N,

    k=mnakk=mnak<ϵ|\sum_{k=m}^{n} a_k| \leq \sum_{k=m}^n |a_k|< \epsilon (By Triangular Inequality)

    Then by Fact 2 again, n=1an\sum_{n=1}^{\infty} a_n  converges.

    Compare: F1.2: convergence of {an}\{a_n\}\Rarr convergence of {an}\{|a_n|\} but not the reverse. 因为不取和则绝对值更易converge,它可以把和两岸的人团结成一股力量(Der Mensch ist ein Seil über einem Abgrunde.)

  • 3.4 Comparison Test: Suppose 0anbn,n0\leq a_n \leq b_n, \forall n \geq  some n0n_0, then
    • 内收 If bn\sum b_n converges, then so does an\sum a_n.
    • 外推 If an\sum a_n diverges, then so does bn\sum b_n.
    • Proof

      (i) WLOG, assume n0=1n_0=1.

      anbnsnasnba_n \leq b_n \Rightarrow s_n^{a} \leq s_n^{b}. Moreover, since bn\sum b_n converges, limsnb\lim s_n^{b} exists as a finite number, then {snb}\{s_n^{b}\} is bounded, denote the bound by MM. Therefore, snasnbMs_n^{a} \leq s_n^{b} \leq M.

      In addition, an0a_n \geq 0, snas_n^{a} is monotone increasing and bounded below by 0. Consequently, {sna}\{s_n^{a}\} is monotonically increasing and bounded, by MCT, limsna\lim s_n^{a} exists as a finite number, and thus an\sum a_n converges.

      (ii) Since an\sum a_n  diverges and {sna}\{ s_n^{a}\}  is monotonically increasing, by MCT, we have limsna=\lim s_n^a=\infty. Then limsnb=\lim s_n^b =\infty and bn\sum b_n diverges.

    • Remark: If an0a_n\ge0, then an\sum a_n either converges to a finite number or an=\sum a_n=\infin.
    • Examples: 常见锚点p-series/geometric series/telescope sum, i.e., 1xp,βn\sum \frac{1}{x^p},\sum \beta^n and (xnxn1)\sum (x_n-x_{n-1}).
      • Hw3 T8 (a): an=n+1na_n=\sqrt{n+1}-\sqrt{n} Back to index

        Multiplying and dividing ana_n by n+1+n\sqrt{n+1}+\sqrt{n}, we find that

        an=1n+1+n>12n+1a_n=\frac{1}{\sqrt{n+1}+\sqrt{n}}>\frac{1}{2\sqrt{n+1}}

        Since 1n\sum \frac{1}{\sqrt{n}}  diverges (p=12<1p=\frac{1}{2}<1), hence an\sum a_n diverges

        Alternatively, since the sum telescopes, the nnth partial sum sn=n+11s_n=\sqrt{n+1}-1, which obviously tends to infinity.

      • Hw3 T8 (b): an=n+1nna_n=\frac{\sqrt{n+1}-\sqrt{n}}{n} Back to index

        Using the same trick above, we find that an=1n[n+1+n]<12n32a_n=\frac{1}{n[\sqrt{n+1}+\sqrt{n}]}<\frac{1}{2 n^{\frac{3}{2}}}.

        Since 1n32\sum \frac{1}{n^{\frac{3}{2}}} converges (p=32>1p=\frac{3}{2}>1), by comparison test, an\sum a_n converges

      • Hw3 T9: Let an0a_n \geq 0, prove that the convergence of an\sum a_n implies the convergence of ann\sum \frac{\sqrt{a_n}}{n} [Hint: Comparison test and aba2+b22|ab|\leq \frac{a^2+b^2}{2}] Back to index

        Note that ann=an×1n(an)2+(1n)22=12(an+1n2)\frac{\sqrt{a_n}}{n}=\sqrt{a_n}\times \frac{1}{n}\leq \frac{(\sqrt{a_n})^2+(\frac{1}{n})^2}{2}=\frac{1}{2}(a_n+\frac{1}{n^2})

        Since an\sum a_n converges, 1n2\sum \frac{1}{n^2} converges, we have (an+1n2)\sum (a_n+\frac{1}{n^2}) converges.

        By comparison test, ann\sum \frac{\sqrt{a_n}}{n} converges

      • HW3 T10 (c): Suppose an>0a_n>0, Sn=a1+...+anS_n=a_1+...+a_n and an\sum a_n diverges. Prove that ansn21sn11sn\frac{a_n}{s_n^2}\leq \frac{1}{s_{n-1}}-\frac{1}{s_n} and deduce that ansn2\sum \frac{a_n}{s_n^2} converges. Back to index

        1sn11sn=snsn1sn1sn=ansn1snansn2\frac{1}{s_{n-1}}-\frac{1}{s_n}=\frac{s_n-s_{n-1}}{s_{n-1}s_n}=\frac{a_n}{s_{n-1}s_n} \geq \frac{a_n}{s_n^2}

        Denote ssn=k=2n(1sk11sk)=1s11snss_n=\sum_{k=2}^{n} (\frac{1}{s_{k-1}}-\frac{1}{s_k})=\frac{1}{s_1}-\frac{1}{s_n}. Hence

        limssn=1s1lim1sn=1s1\lim ss_n=\frac{1}{s_1}-\lim \frac{1}{s_n}=\frac{1}{s_1}

        Therefore, (1sn11sn)\sum (\frac{1}{s_{n-1}}-\frac{1}{s_n}) converges. By comparison test, ansn2\sum \frac{a_n}{s_n^2} converges.

      • Hw3 T10 (d): Construct a sequence {an}\{a_n\} such that an\sum a_n diverges but an1+nan\sum \frac{a_n}{1+na_n}

        Let an=1n2a_n=\frac{1}{n^2} if nn is not a perfect square and an=1na_n=\frac{1}{\sqrt{n}} if nn is a perfect square (the bad guy). Back to index

        ann=k21k=k=11k=\sum a_n \geq \sum_{n=k^2} \frac{1}{k}=\sum_{k=1}^{\infty} \frac{1}{k}=\infty, hence an\sum a_n  diverges

        But an1+nan=nk21n2+n3+n=k21k+k2n=11n2+n3+k=11k+k2\sum \frac{a_n}{1+na_n} = \sum_{n\neq k^2} \frac{1}{n^2+n^3}+\sum_{n=k^2} \frac{1}{k+k^2} \leq \sum_{n=1}^{\infty} \frac{1}{n^2+n^3}+\sum_{k=1}^{\infty}\frac{1}{k+k^2}

        Both n=11n2+n3\sum_{n=1}^{\infty} \frac{1}{n^2+n^3} and k=11k+k2\sum_{k=1}^{\infty} \frac{1}{k+k^2} converges, hence an1+nan\sum \frac{a_n}{1+na_n} converges.

        @Cindy Zeng 你觉得这个地方的rearrangement 可能会有问题嘛,就把能开平方数和非平方数分别求和。 虽然这里ana_n 都是正数,如果有convergence也有absolute convergence, 但有点预设了你要证明的东西的感觉。

        @The Grand Inquisitor 我觉得没有问题哇,只要找到一个反例即可,如果这个反例确实满足原序列diverge 新序列converge 那不就可以了?不过为什么要提absolute convergence?

        @Cindy Zeng 我担心的是如果原序列不是absolute convergent的话,rearrangement不一定等于原序列的和。We are too young to know too much

        @The Grand Inquisitor I don’t think this is a rearrangement though. Its order is preserved, it’s just that you group them when you do addition. Is it? @The Grand Inquisitor

        @The Grand Inquisitor 天,老师最后讲rearrangement 的那个proof的时候,最后蓝笔才是正确版本对吗?相当于之前的黑笔版本就不要了?

      • Hw3 T11 (b): Back to index

        2rn2rnrn+1=rnrn+1+rn+1+rn2rnrn+1=an+rn+1+rn2rnrn+1>an2r_n-2\sqrt{r_nr_{n+1}}=r_n\colorbox{greenyellow}{$-r_{n+1}+r_{n+1}$}+r_n-2\sqrt{r_nr_{n+1}}=a_n+r_{n+1}+r_n-2\sqrt{r_nr_{n+1}}>a_n. Divide both sides by rn\sqrt{r_n} we have the desired inequality. Since k=1n(rkrk+1)=r1rn+1\sum_{k=1}^n(\sqrt{r_k}-\sqrt{r_{k+1}})=\sqrt{r_1}-\sqrt{r_{n+1}} (telescoping), and limrn=l,k=1n(rkrk+1)\lim r_n=l,\sum_{k=1}^n(\sqrt{r_k}-\sqrt{r_{k+1}}) converges. By comparison test, … converges.

  • 3.5 Limiting Comparison Test: Suppose an,bn0,na_n,b_n \geq 0, \forall n \geq some n0n_0, if limnanbn=l\lim_{n\to \infty} \frac{a_n}{b_n}=l
    • 辅车相依,唇亡齿寒 0<l<0<l<\infty: an\sum a_n and bn\sum b_n converge/diverge at the same time
    • l=0l=0: convergence of bn\sum b_n implies convergence of an\sum a_n and divergence of an\sum a_n implies divergence of bn\sum b_n
    • l=l=\infty: convergence of an\sum a_n implies convergence of bn\sum b_n and divergence of bn\sum b_n implies the divergence of an\sum a_n
    • Proof of (1) 白矛纯束:

      ϵ>0,N,s.t.nN:lϵ<anbn<l+ϵ\forall \epsilon>0, \exist N, s.t. \forall n \geq N: l-\epsilon<\frac{a_n}{b_n}<l+\epsilon

      Take ϵ=l2\epsilon=\frac{l}{2}, then we have l2<anbn<3l2,nN\frac{l}{2}<\frac{a_n}{b_n}<\frac{3l}{2}, \forall n \geq N

      Thus l2bn<an<3l2bn,nN\frac{l}{2}b_n<a_n<\frac{3l}{2}b_n, \forall n \geq N. Then by Comparison Test, if an\sum a_n converges, then so does l2bn\sum\frac{l}{2}b_n, and so does bn\sum b_n. Otherwise, if an\sum a_n diverges, then so does 3l2bn\sum \frac{3l}{2} b_n, and so does bn\sum b_n

    • Proof of (2):

      anbn0ϵ>0,N s.t. if nN,anbn<ϵan<ϵbn.\because \frac{a_n}{b_n}\rarr0\therefore\forall \epsilon>0,\exists N~s.t.~if~n\ge N,\frac{a_n}{b_n}<\epsilon\Rarr a_n<\epsilon b_n. Take ϵ=10anbn(2)\epsilon=1\Rarr0\le a_n\le b_n\Rarr(2) follows from 3.4 Comparison Test.

    • Examples
      • Hw3 T10 (a): Suppose an>0a_n>0, Sn=a1+...+anS_n=a_1+...+a_n and an\sum a_n diverges. Prove that an1+an\sum \frac{a_n}{1+a_n} diverges. Back to index

        Tian:

        Proof by contradiction: Suppose an1+an\sum \frac{a_n}{1+a_n} converges, then by divergence test, liman1+an=0\lim \frac{a_n}{1+a_n}=0. Notice that an1+an=111+an\frac{a_n}{1+a_n}=1-\frac{1}{1+a_n}, hence liman1+an=0\lim \frac{a_n}{1+a_n}=0 implies that lim11+an=1\lim \frac{1}{1+a_n}=1.

        Take the ratio between ana_n and an1+an\frac{a_n}{1+a_n} : an1+anan=11+an\frac{\frac{a_n}{1+a_n}}{a_n}=\frac{}{}\frac1{1+a_n}{}.

        Therefore, liman1+anan=lim11+an=1\lim \frac{\frac{a_n}{1+a_n}}{a_n}=\lim \frac{1}{1+a_n}=1 is a constant, an1+an\sum \frac{a_n}{1+a_n} and an\sum a_n should diverge simultaneously, contradiction.

        Cind:

        Proof by contradiction:

        Suppose an1+an\sum\frac{a_n}{1+a_n} conv fact 3.1liman1+an=0=isolate anlim1+an1+an11+anlim11+an=1lim1+an=1liman=0.\xRightarrow{fact~3.1}\lim\frac{a_n}{1+a_n}=0\overset{\text{isolate an}}{=}\lim\frac{1+a_n}{1+a_n}-\frac{1}{1+a_n}\Rarr\lim\frac{1}{1+a_n}=1\Rarr\lim1+a_n=1\Rarr\lim a_n=0.

        anan>0an1+an\because a_n\overset{a_n>0}{\ge}\frac{a_n}{1+a_n}\therefore Comparison test (fact 3.4) doesn’t work.

        Use limiting comparison test (fact 3.5) limanan/(1+an)=lim(1+an)=1(0,).\lim\frac{a_n}{a_n/(1+a_n)}=\lim(1+a_n)=1\in(0,\infty). So an\sum a_n must converge with an1+an\sum\frac{a_n}{1+a_n}, which contradicts with the setup.

  • 3.6 Suppose an0,n1a_n \geq 0, \forall n\geq 1. Then an\sum a_n  converges iff {sn}\{s_n\} is bounded
    • @The Grand Inquisitor 天,为啥我突然觉得这里有点问题,sn=(1)ns_n=(-1)^n 咋办。我的意思是如果这条fact成立那不就意味着 sn convergent \Lrarr sn bounded, which is not true.?
    • 这里的前提是an0a_n\geq0, 所以sns_n是单增的,单增的话convergence和boundedness就是可以互推的 @Cindy Zeng 哦对,我傻了 @The Grand Inquisitor
  • 3.7 Integral Test 平行宇宙: Suppose an0a_n \geq 0, {an}\{a_n\} is decreasing, and there exists a continuous and decreasing function f(x)f(x) defined on [1,)[1,\infty) s.t. an=f(n)a_n=f(n). Then an&1f(x)dx\sum a_n \& \int_{1}^{\infty} f(x)dx converge or diverge at the same time.
    • Proof of Case 1: 1f(x)dx\int_{1}^{\infty} f(x)dx converges.

      Since an0a_n \geq 0, sns_n is increasing, to prove an\sum a_n converges, we only need to show {sn}\{s_n\} is bounded.

      sn=k=1nana1+12f(x)dx+...+n1nf(x)dx=a1+1nf(x)dxs_n=\sum_{k=1}^{n}a_n\leq a_1+\int_{1}^{2} f(x) dx+...+\int_{n-1}^n f(x)dx=a_1+\int_{1}^{n} f(x)dx

      Since 1f(x)dx=limk1kf(x)dx\int_{1}^{\infty}f(x)dx=\lim_{k \to \infty} \int_{1}^{k} f(x)dx  exists & finite, we have 1nf(x)dx\int_{1}^{n} f(x)dx is bounded(Convergence implies boundedness). Therefore, sns_n is bounded.

    • Proof of Case 2: 1f(x)dx\int_{1}^{\infty} f(x)dx diverges

      Similarly, to prove an\sum a_n diverges, we only need to show {sn}\{s_n\} is unbounded.

      sn=k=1nan12f(x)dx+...+nn+1f(x)dx=1n+1f(x)dxs_n =\sum_{k=1}^{n} a_n \geq \int_{1}^2 f(x)dx+...+\int_{n}^{n+1} f(x)dx= \int_{1}^{n+1} f(x)dx

      The LHS approaches infinity as n goes to infinity, hence {sn}\{s_n\} is unbounded.

    • Example 9: p-series n=11np\sum_{n=1}^\infty\frac1{n^p}, converges if p>1,p>1, diverges if 0<p10<p\le 1 (or, p0p\le 0). Back to index

      Let f(x)=1xpf(x)=\frac1{x^p}, which decreases on [1,)[1,\infin), and f(n)=1np=an.f(n)=\frac1{n^p}=a_n.

      11xpdx=1xpdx={xp+1p+11p1lnx1p=111xpdx={p<11p1p>1p=1\int_1^\infty\frac1{x^p}dx=\int_1^\infty x^{-p}dx=\begin{cases} \frac{x^{-p+1}}{-p+1}\vert^\infty_1 & p\ne 1 \\ lnx\vert^\infty_1 & p=1\end{cases}\Rarr\int_1^\infty\frac1{x^p}dx=\begin{cases} \infty & p<1\\ \frac1{p-1} & p>1\\ \infty & p=1\end{cases}

      \Rarr By Integral Test, n=11np\sum_{n=1}^\infty\frac1{n^p} converges if p>1,p>1, diverges if p1p\le 1.

  • 3.8 Root Test: Let α=lim supnan1n\alpha=\limsup_{n\to \infty} |a_n|^{\frac{1}{n}}
    • If α<1\alpha <1, then an\sum |a_n| converges

      Take β(α,1)\beta \in (\alpha,1), Then N,s.t.an1n<β,nN\exist N, s.t. |a_n|^{\frac{1}{n}}< \beta, \forall n \geq N

      an<βn,nN\Rightarrow |a_n| < \beta ^n, \forall n \geq N.

      Since βn\sum \beta^n  as a geometric series converges, then by comparison test, an\sum |a_n| converges.

    • If α>1\alpha >1, then an\sum |a_n| diverges

      Since lim supnan1n=α\limsup_{n \to \infty} |a_n|^{\frac{1}{n}}= \alpha, then there exists a subsequence {ank1nk}k=1\{ |a_{n_k}|^{\frac{1}{n_k}}\}|_{k=1}^{\infty} such that limkank1nk=α\lim_{k \to \infty} |a_{n_k}|^{\frac{1}{n_k}}=\alpha (F2.3.2). Consequently, take ϵ=α12\epsilon=\frac{\alpha-1}{2} (so small that αϵ>1\alpha-\epsilon>1), there exists KK such that

      αϵ<ank1nk<a+ϵ\alpha-\epsilon < |a_{n_k}|^{\frac{1}{n_k}}<a+\epsilon

      as long as kKk \geq K. Hence ank>(αϵ)nk,kK|a_{n_k}|> (\alpha-\epsilon)^{n_k}, \forall k \geq K.

      But αϵ>1\alpha-\epsilon>1, hence limk(αϵ)nk=ankliman0div testan\lim_{k \to \infty} (\alpha-\epsilon)^{n_k} =\infty\Rarr|a_{n_k}|\rarr\infty\Rarr\lim |a_n|\ne 0\overset{\text{div test}}{\Rarr}\sum |a_n| diverges.

    • If α=1\alpha=1, inconclusive

      Counterexample: 1n\sum\frac1n diverges, 1n2\sum \frac1{n^2} converges. Yet both of their α=limnan1n=1\alpha=\varlimsup_{n\to \infty} |a_n|^{\frac{1}{n}}=1.

    • Example (Hw4 T2) Back to index

      3个root test, 1个ratio test(见到阶乘时的条件反射)

  • 3.9 Ratio Test an+1an|\frac{a_{n+1}}{a_n}|
    • 3.9.1 If lim supnan+1an<1\limsup_{n \to \infty} |\frac{a_{n+1}}{a_n}|<1, then an\sum |a_n| converges.

      We can find α<1\alpha<1 and NN so that an+1an<α|\frac{a_{n+1}}{a_n}|<\alpha, nN\forall n \geq N

      an+1<αan\Rightarrow |a_{n+1}|<\alpha |a_n|

      Take n=Nn=N, aN+1<αaN|a_{N+1}|<\alpha |a_N|

      Take n=N+1n=N+1, aN+2<αaN+1<α2aN|a_{N+2}|<\alpha|a_{N+1}|< \alpha^2 |a_{N}|

      By induction, we have aN+p<αaN+p1<αpaN|a_{N+p}|< \alpha |a_{N+p-1}|< \alpha^p |a_{N}|

      Then p=1aN+p<(p=1αp)aN<\sum_{p=1}^{\infty}|a_{N+p}|< (\sum_{p=1}^{\infty} \alpha^p)|a_N|< \infty (infinite geometric series, which is finite, times a finite number).

      Thus n=N+1an\sum_{n=N+1}^{\infty} |a_n| converges (by MCT), which implies an\sum |a_n| converges (by Fact 3.1).

    • 3.9.2 If n0,s.t.nn0,an+1an1,\exist\, n_0,s.t. \forall n \geq n_0,|\frac{a_{n+1}}{a_n}| \geq 1,  then an\sum a_n diverges.

      This means an|a_n| is increasing. By MCT, liman=l\lim |a_n|=l exists (may be \infty)

      But a1>1|a_1|>1, then liman=l0\lim |a_n|=l \neq 0. By divergence test, an\sum a_n diverges

    • 3.9.3 If lim infan+1an>1\liminf |\frac{a_{n+1}}{a_n}| >1, then an\sum a_n diverges.

      Follows from 3.9.2, by F2.3.1.

  • Theorem: Root Test is more general than Ratio Test
    • lim infan+1anlim infan1nlim supan1nlim supan+1an\liminf |\frac{a_{n+1}}{a_n}| \leq \liminf |a_n|^{\frac{1}{n}} \leq \limsup |a_n|^{\frac{1}{n}} \leq \limsup |\frac{a_{n+1}}{a_n}|
    • Prove lim infan+1anlim infan1n\liminf |\frac{a_{n+1}}{a_n}|\leq \liminf |a_n|^{\frac{1}{n}}

      Let β=lim infan+1an0\beta=\liminf|\frac{a_{n+1}}{a_n}| \geq 0

      If β=0\beta=0, then lim infan+1an=0lim infan1n\liminf |\frac{a_{n+1}}{a_n}|=0 \leq \liminf |a_n|^{\frac{1}{n}}, done.

      If 0<β<0<\beta<\infty, then take γ(0,β)\gamma\in (0,\beta), then there exists NN,such that nN\forall n \geq N,

      an+1an>γan+1γanaN+pγpaN|\frac{a_{n+1}}{a_n}| >\gamma \Rightarrow |a_{n+1}|\geq \gamma |a_n| \Rightarrow|a_{N+p}| \geq \gamma^{p} |a_N|

      Hence akγkNaN,kN+1|a_{k} |\geq \gamma^{k-N}|a_N|, \,\,\forall k \geq N+1

      Thus ak1k>γkNkaN1k|a_k|^{\frac{1}{k}} > \gamma^{\frac{k-N}{k}}a_{N}^{\frac{1}{k}}. Take liminf on both sides (RHS has limits)

      lim infak1kγ\Rightarrow \liminf |a_k|^{\frac{1}{k}} \geq \gamma

      Since this fixed γ\gamma is an arbitrary element from (0,β)(0,\beta), we have

      lim infak1ksup{(0,β)}=β\liminf |a_k|^{\frac{1}{k}} \geq \sup\{(0,\beta)\}=\beta

    • Prove lim supan1nlim supan+1an\limsup|a_n|^{\frac{1}{n}} \leq \limsup |\frac{a_{n+1}}{a_n}|

      Let α=lim supan+1an\alpha=\limsup |\frac{a_{n+1}}{a_n}|.

      If α=,\alpha=\infty,  then lim supan1n=lim supan+1an\limsup |a_n|^{\frac{1}{n}} \leq \infty=\limsup |\frac{a_{n+1}}{a_n}|.

      If α<\alpha< \infty, then take γ>α\gamma> \alpha, then there exists NN, such that nN\forall n \geq N,

      an+1an<γan+1<γanaN+p<γpaN|\frac{a_{n+1}}{a_n}|<\gamma \Rightarrow |a_{n+1}|<\gamma |a_n| \Rightarrow |a_{N+p} |<\gamma^{p} |a_{N}|

      Thus nN,an1n<γnNnaN1n\forall n \geq N, |a_{n}|^{\frac{1}{n}}<\gamma^{\frac{n-N}{n}}|a_N|^{\frac{1}{n}}. Take limsup on both sides,

      lim supan1nγ\limsup |a_n|^{\frac{1}{n}} \leq \gamma

      Since this fixed γ\gamma is an arbitrary element in (α,)(\alpha, \infty), we have

      lim supan1ninf(α,)=α\limsup |a_n|^{\frac{1}{n}} \leq \inf{(\alpha,\infty)} =\alpha

    • Corollary, if liman+1an\lim |\frac{a_{n+1}}{a_n}| exists, then liman1n\lim |a_n|^{\frac{1}{n}} also exists and equals liman+1an\lim |\frac{a_{n+1}}{a_n}|

      Proof: Squeezing Principle

    • Example 10: Try root test and ratio test on series: 12+13+122+132+...\frac12+\frac13+\frac1{2^2}+\frac1{3^2}+...Back to index

      Root test:

      a2k1=12k,k0,a2k=13k,k1.a_{2k-1}=\frac1{2^k},\forall k\ge0,a_{2k}=\frac1{3^k},\forall k\ge 1. Want to know liman1n.\varlimsup |a_n|^\frac1n.

      Let liman1n=supE. an1n={12k2k1,n=2k11213k2k,n=2k=13E=12,13,supE=12<1.\varlimsup |a_n|^\frac1n=\sup E.~|a_n|^\frac1n=\begin{cases} \frac1{2^\frac{k}{2k-1}}, & n=2k-1 & \rarr\frac1{\sqrt2}\\ \frac1{3^\frac{k}{2k}},& n=2k &=\frac1{\sqrt3} \end{cases}\Rarr E={\frac1{\sqrt2},\frac1{\sqrt3}},\sup E=\frac1{\sqrt2}<1.

      By root test the series converges absolutely.

      Ratio test:

      Want to know liman+1an.\varlimsup \frac{a_{n+1}}{a_n}.

      an+1an={(23)k,n=2k1(32)k12,n=2kliman+1an=,liman+1an=0\frac{a_{n+1}}{a_n}=\begin{cases} (\frac23)^k, & n=2k-1\\ (\frac32)^k\cdot\frac12, & n=2k \end{cases}\Rarr \varlimsup \frac{a_{n+1}}{a_n}=\infty,\varliminf \frac{a_{n+1}}{a_n}=0

      Ratio test fails.

      Root test prevails.

The Magic Number e

4. Basic Point-Set Topology

Topology, no trick, all conceptual problems

5. Compact Sets

6. Connected Sets

7. Limit of Functions

8. Continuity of Functions

9. Discontinuity of Functions

10. Differentiation

Back to index

11. Riemann Integral

@The Grand Inquisitor 哈哈小天,我在听最后两节课的录播,他说math research is much easier, you just need a pen and paper. 突然觉得做数学应该和做哲学很像吧。都是闭门造车。除非你做马克思那样的哲学哈哈哈。

@The Grand Inquisitor 天,最后一节课的内容是不是完全不会考

@Cindy Zeng 我刚在搞3040,还没开始看,最后一节你先别听吧,我当时听完倒数第二节感觉他已经把那三个问题回答的差不多了,我听完再告诉你有没有啥重点,不过听听课确实能找回一些rhythm

@The Grand Inquisitor 天,HW 13 P1 你作业里写的方法是绝对严谨的吧?感觉你和TA意思一样,就是在取xR,yRx_R,y_R的时候多用了一个消除中间的巧思……考试写你的方法应该木有问题吧?

@Cindy Zeng 嗯,他中间用那个εδ\varepsilon-\delta没有必要,直接用 limp0i=1nf(xR)R\lim_{|p|\to 0} \sum_{i=1}^n f(x_R)|R| 来算积分就行,只是这里的Riemann sum 搞出来加减相消后是f(xn)xnf(x0)x0f(x_n)x_n-f(x_0)x_0p|p|无关,稍微显得有点不自然。今天不知道咋回事,用notion又开始卡了😩

@The Grand Inquisitor 我也卡。而且/eq自动给我搞成block equation了不知道怎么调回inline

@Cindy Zeng 用shift+command+E可以直接快捷键打开inline equation,但整体打字延迟还是很高,尤其是中文

@The Grand Inquisitor 可能真的是文件太大了,有嗲担心他随时崩掉。另外我看了HW13最后一题,似乎倒数第二页写完也就只需要下个结论了?也许那张失效的图不是作业答案?

趁你睡觉的时候押题一波:

小题:

  1. HW8P6 记得quiz 也出过。

大题(期中前*2):

  1. Qz3P1 Construct a bad sequence (of indices) from a NOT-Cauchy sequence.

大题(期中后*9):

  1. HW13P4 积分域交集零测则积分可加和。
  1. HW13P5 利用continuity反证的经典,sharpened IVT.
  1. HW12P2 Continuity证明,定义在闭区间上的连续函数零测的证明(这在原本的简单对角线的情况下不需要但是出一道变式,把平面变成空间,把f=1变成some continuous function就可以融合HW12P5,能考到很多东西。我预感他会有变式,后来他用的动词都不是“is”而是“related to” HW problems了。那道新题应该也是某种变化比较大的变式。
  1. HW11P5 The first set of sufficient conditions that makes ff an open mapping: 1) ff is C1C^1 smooth, 2) Jacobian f(x)f'(x) is non-singular x\forall x. “Extremely easy exercise, but a signicant result.”
  1. HW8P5 NUS problem. “Try this yourself to see how much time you will take to come up with a correct and pretty proof!” It is pretty indeed.

次有可能考的:

  1. HW12P1 挺综合的,把Riemann Integrability 原始定义的那些概念都考到了

@The Grand Inquisitor 天,Finally那段他到底在讲什么,我看了半天都没看懂。。this result 指代的是哪个?不需要(f1)(f^{-1})' non singular 吗。

@Cindy Zeng 我上午在高铁上,这边打起来依然很卡,我用overleaf给你押题好了